Mathematics 375 – Probability and Statistics I Solutions – Problem Set 9 December 7, 2005

5.22/ a) The marginal densities are f_{1}(y_{1}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{2} = 2y_{1} and similarly f_{2}(y_{2}) = ∫_{0}^{1} 4y_{1}y_{2} dy_{1} = 2y_{2}. Looking ahead to the next section, note that these satisfy the criterion of Theorem 5.4. So Y_{1} and Y_{2} are independent.

b) To compute the denominator in the conditional probability, we can use the marginal density from part a:

P (Y_{1} ≤ 1/2|Y_{2} ≥ 3/4) = \frac{P (Y_{1} ≤ 1/2, Y_{2} ≥ 3/4)}{P (Y_{2} ≥ 3/4)}

= \frac{∫_{0}^{1/2} ∫_{3/4}^{1} 4y_{1}y_{2} dy_{2} dy_{2}}{∫_{3/4}^{1} 2y_{2} dy_{2}}

= \frac{∫_{0}^{1/2} 2y_{1} dy_{1} ∫_{3/4}^{1} 2y_{2} dy_{2}}{∫_{3/4}^{1} 2y_{2} dy_{2}}

= ∫_{0}^{1/2} 2y_{1} dy_{1}

= 1/4

(The cancellation works this way because of independence.)

5.27/ The region where the density is nonzero here is the interior of the triangle with vertices at (0, 1), (1, 0), (0, −1). As always, to find the marginal density with respect to one of the variables we compute the “inside” integral of the iterated double integral of the joint density with the “other variable” on the inside. The bottom edge of this triangle is the line y_{2} = y_{1} − 1 and the top edge is y_{2} = 1 − y_{1}. a) Hence to find f_{1}(y_{1}) we compute for 0 ≤ y_{1} ≤ 1:

∫_{y_{1}−1}^{1−y_{1}} 30y_{1}y_{2}^{2} dy_{2} = 20y_{1}(1 − y_{1})^{3}

(Outside 0 ≤ y_{1} ≤ 1, the marginal density is zero.) From Chapter 4, we recognize this as the form for a Beta pdf with α = 2, β = 4.

b) The marginal density f_{2}(y_{2}) is not as nice because the right boundary curve of the triangle changes depending on the value of y_{2}. If 0 ≤ y_{2} ≤ 1, then f_{2}(y_{2}) = ∫_{0}^{1−y_{2}} 30y_{1}y_{2}^{2} dy_{1} = 15y_{2}^{2}(1 − y_{2})^{2}. On the other hand, If −1 ≤ y_{2} ≤ 0, then f_{2}(y_{2}) = ∫_{0}^{y_{2}+1} 30y_{1}y_{2}^{2} dy_{1} = 15y_{2}^{2}(y_{2} + 1)^{2}.

5.31/We take the factor depending only on y_{2} outside the integral with respect to y_{1}, then change variables to u = y_{1}/2:

f_{2}(y_{2}) = ∫_{0}^{∞} \frac{1}{8}y_{1}e^{−(y_{1}+y_{2})/2} dy_{1}

= \frac{1}{2} ∫_{0}^{∞} ue^{−u} du · e^{−y_{2}/2}

= \frac{1}{2}Γ(2)e^{−y_{2}/2}

= \frac{1}{2}e^{−y_{2}/2}

if y_{2} > 0, and zero otherwise. This gives the marginal density for Y_{2}. Then the desired probability is found by integrating

∫_{2}^{∞} f_{2}(y_{2}) dy_{2} = ∫_{2}^{∞} \frac{1}{2}e^{−y_{2}/2} dy_{2} = 1/e.

(Note that the problem should say the life of the component is in 100’s of hours to match the previous problem this one builds on. So the lower limit of integration is 2 rather than 200.)

5.32 a) We claim

f(y_{1}, y_{2}) = \{ \begin{matrix}1/y_{1} & if 0 < y_{1} ≤ 1, 0 ≤ y_{2} ≤ y_{1} \\ 0 & otherwise \end{matrix}

The region where this is nonzero is the interior of the triangle with vertices (0, 0), (1, 0), (1, 1). To convince ourselves this is correct, we can compute the marginal densities and see that they match what the problem says. First, if 0 < y_{1} ≤ 1, f_{1}(y_{1}) = ∫_{0}^{y_{1}} 1/y_{1} dy_{2} = y_{2}/y_{1}|_{0}^{y_{1}} = 1. (Outside 0 < y_{1} ≤ 1 we get zero.) Hence we do get the uniform density on (0, 1). Second, If we fix a value for y_{1}, then the “conditional” density f(y_{2}|Y_{1} = y_{1}) =

\{ \begin{matrix}1/y_{1} & if 0 ≤ y_{2} ≤ y_{1} \\ 0 & otherwise \end{matrix} which is also correct for the uniform distribution on (0, y_{1}).

b) P (Y_{2} ≥ 1/4|Y_{1} = 1/2) = ∫_{1/4}^{1/2} 2 dy_{2} = 1/2.

5.39/ Not independent. For instance P (Y_{1} = 0, Y_{2} = 0) = 4/9 ≠ P (Y_{1} = 0)P (Y_{2} = 0) = 16/81. (We compute the marginal distributions to find P (Y_{1} = 0) = 1/9+2/9+1/9 = 4/9 and similarly for P (Y_{2} = 0).

5.46/ No – Y_{1}, Y_{2} cannot be independent when the interval of y_{2} values where the density is nonzero depends on y_{1}. (You can also see this directly by computing f_{1}(y_{1}) and f_{2}(y_{2}). The region where the joint density is nonzero is the triangle with vertices at (0, 0), (2, 0), (2, 1). So

f_{1}(y_{1}) = ∫_{0}^{y_{1}/2} 1 dy_{2} = \frac{y_{1}}{2}

(when 0 ≤ y_{1} ≤ 2; 0 otherwise). Similarly,

f_{2}(y_{2}) = ∫_{2y_{2}}^{2} 1 dy_{1} = 2 − 2y_{2}

(when 0 ≤ y_{2} ≤ 1; 0 otherwise). The product f_{1}(y_{1})f_{2}(y_{2}) is not the same as f(y_{1}, y_{2}).)

5.64/ When the function depends on only one of Y_{i}, we can use the marginal densities computed in Exercise 5.22 above. We will use that method in parts a and b. For functions depending on both variables, as in part c, we integrate against the joint density to find the expected value.

a) E(Y_{1}) = ∫_{0}^{1} 2y_{1}^{2} dy_{1} = 2/3.

b) E(Y_{1}^{2}) = ∫_{0}^{1} 2y_{1}^{3} dy_{1} − (2/3)^{2} = 1/2 − 4/9 = 1/18.

c) E(Y_{1} − Y_{2}) = ∫_{0}^{1} ∫_{0}^{1}(y_{1} − y_{2}) · 4y_{1}y_{2} dy_{2} dy_{1} = 0. (The last one can also be seen using the linearity E(Y_{1} − Y_{2}) = E(Y_{1}) − E(Y_{2}) then noting symmetry of the joint density. The two terms must cancel.)

5/70. Instead of splitting up the expected value or computing marginal densities, the fastest way to compute this is simply to take

E(Y_{1} − Y_{2}) = ∫_{0}^{1} ∫_{0}^{y_{1}}(y_{1} − y_{2})\frac{1}{y_{1}} dy_{2} dy_{1} = ∫_{0}^{1} y_{1}/2 dy_{1} = 1/4.

5.77/ Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) − E(Y_{1})E(Y_{2}) = 0 since we know Y_{1} and Y_{2} are independent. (See Exercise 5.22 above and Theorem 5.11.)

5.87/ By linearity of expected values, E(3Y_{1} + 4Y_{1} − 6Y_{3}) = 3E(Y_{1}) + 4E(Y_{2}) − 6E(Y_{3}) = 6 − 4 − 24 = −22. We use Theorem 5.12b to find

V (3Y_{1} + 4Y_{1} − 6Y_{3}) = 9V (Y_{1}) + 16V (Y_{2}) + 36V (Y_{3})
+ 24Cov(Y_{1}, Y_{2}) − 18Cov(Y_{1}, Y_{3}) − 48Cov(Y_{2}, Y_{3})

= 36 + 96 + 288 + 24 + 36 + 0 = 480.

5.92/ By computing marginal densities here, we see Y_{1} has a Gamma distribution with α = 2 and β = 1. Y_{2} is exponentially distributed. So E(Y_{1}) = 2, E(Y_{2}) = 1, and E(Y_{1} − Y_{2}) = 1. An alternate way to see this: integrate against the joint density:

E(Y_{1} − Y_{2}) = ∫_{0}^{∞} ∫_{0}^{y_{1}}(y_{1} − y_{2})e^{−y_{1}} dy_{2} dy_{1}

= \frac{1}{2} ∫_{0}^{∞} y_{1}^{2}e^{−y_{1}} dy_{1}

= \frac{1}{2}Γ(3)

= 1 Then by Theorem 5.12 b,

V (Y_{1} − Y_{2}) = V (Y_{1}) + V (Y_{2}) − 2Cov(Y_{1}, Y_{2})
= 2 + 1 − 2(E(Y_{1}Y_{2}) − 2 · 1)
= 7 − 2 ∫_{0}^{∞} ∫_{0}^{y_{1}} y_{1}y_{2}e^{−y_{1}} dy_{2} dy_{1}
= 7 − ∫_{0}^{∞} y_{1}^{3}e^{−y_{1}} dy_{1}
= 7 − Γ(4)
= 7 − 6 = 1.

It is unlikely that a customer spends 4 minutes or more at the window, since 4 = μ + 3σ (three standard deviations above mean).

6.33/ U = Y_{1}/2 + Y_{2}/2 where each Y_{i} is exponentially distributed and Y_{1}, Y_{2} are independent. Hence by Theorem 6.2, the mgf for U is

m_{U}(t) = m_{Y_{1}}(t/2) · m_{Y_{2}}(t/2) = \frac{1}{(1 − t/2)^{2}} .

Comparing with the table inside the back cover of the text, we see that U has a Gamma distribution with α = 2 and β = 1/2.

6.34/ Y_{i} are standard normal, so by Discussion 5 (or Example 6.11) Y_{i}^{2} are χ^{2} with ν = 1. They are independent as well. So the mgf for U is

m_{U}(t) = \frac{1}{\sqrt{1 − 2t}} · \frac{1}{\sqrt{1 − 2t}} = \frac{1}{1 − 2t}

Then from the table, U = Y_{1}^{2} + Y_{2}^{2} has a χ^{2} distribution with ν = 2. Therefore f_{U}(u) = \frac{1}{2}e^{−u/2}. (Note: this is also same as an exponential pdf with β = 2.)

6.36/ Since Y_{1} is the capacity of the elevator and Y_{2} is the load, the elevator is overloaded when Y_{2} − Y_{1} > 0. Using the result from Discussion 5, B, U = Y_{2} − Y_{1} is normally distributed with μ_{U} = −1000 and σ_{U}^{2} = 300^{2} + 400^{2} = 500^{2}. The probability that the elevator is overloaded is found by standardizing and using the table for the standard normal

P (Y_{2} − Y_{1} > 0) = P (\frac{(Y_{2} − Y_{1}) + 1000}{500} > 2) = P (Z > 2) = .0228.